Square Root of a number

Dear all
How can i get the square root of a number.for examle 4 from the number 16,as
16 = 4*4.
Best Regards
Bunty

Hii
To get the Square root: use the arithmetic Function SQRT( ).
Sample code: try...
DATA: V1 TYPE I VALUE 100.
DATA : V_ROOT TYPE I.
V_ROOT = SQRT( V1 ).
WRITE:/ 'square root = ', v_root.
<b>reward if Helpful.</b>

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Hi,
Is there any one to make a program for me in Turbo
C++ for Dos, which can calculate the square root of
any number without using the sqrt( ) or any ready
made functions.
The program should calculate the s.root of the number
by a formula or procedure defined by the user
(programmer).
Thanks.This is a Java forum!
If you want Java help:
1. Start your own thread.
2. Use code tags (above posting box) if you post code.
3. No one will write the program for you. We will help by answering your questions and giving advice on how to fix problems in code you wrote.
4. The formula you need to implement is given above by dizzy.

Problems with square root approximations with loops program

i'm having some trouble with this program, this loop stuff is confusing me and i know i'm not doing this correctly at all. the expected values in the tester are not matching up with the output. i have tried many variations of the loop in this code even modifying the i parameter in the loop which i guess is considered bad form. nothing seems to work...
here is what i have for my solution class:
/** A class that takes the inputted number by the tester and squares it, and
* loops guesses when the nextGuess() method is called. The epsilon value is
* also inputted by the user, and when the most recent guess returns a value
* <= epsilon, then the hasMoreGuesses() method should return false.
public class RootApproximator
   /** Takes the inputted values from the tester to construct a RootApproximator.
    * @param val the value of the number to be squared and guessed.
    * @param eps the gap in which the approximation is considered acceptable.
     public RootApproximator(double val, double eps)
          value = val;
          square = Math.sqrt(val);
          epsilon = eps;
   /** Uses the algorithm where 1 is the first initial guess of the
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    * @return increasingly closer guesses as the method is continually used.
   public double nextGuess()
         final int TRIES = 10000;
         double guess = 1;
          for (double i = 1; i < TRIES; i++)
               double temp = value / guess;
               guess = (guess + temp) / 2.0;
          return guess;
   /** Determines if there are more guesses left if the difference
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    * epsilon.
    * @return the value of the condition.
   public boolean hasMoreGuesses()
          return (square - guess <= epsilon);
   private double square;
   private double value;
   private double epsilon;
   private double guess;
here is the tester:
public class RootApproximatorTester
   public static void main(String[] args)
      double a = 100;
      double epsilon = 1;
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and here is the output:
10.0
Expected: 1 // not sure why this should be 1, perhaps because it is the first guess.
10.0
Expected: 50.5 // (100 + 1) / 2, average of the inputted value and the first guess.
true
Expected: true
i'm new to java this is my first java course and this stuff is frustrating. i'm really clueless as to what to do next, if anyone could please give me some helpful advice i would really appreciate it. thank you all.

i'm new to java this is my first java course and this
stuff is frustrating. i'm really clueless as to what
to do nextMaybe it's because you don't have a strategy for what the program is supposed to do? To me it looks like a numerical scheme for finding the squareroot of a number.
Say the number you want to squarerroot is called value and that you have an approximation called guess. How do you determine whether guess is good enought?
Well in hasMoreGuesses you check whether,
(abs(value-guess*guess) < epsilon)
The above decides if guess is within epsilon of being the squareroot of value.
When you calculate the next guess in nextGuess why do you loop so many times? Aren't you supposed to make just one new guess like,
guess = (guess + value/guess)/2.0
The above generates a new guess based on the fact that guess and value/guess must be on each side of value so that the average of them must be closer too value.
Now you can put the two together to a complete algoritm like,
while (hasMoreGuesses()) {
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}In each iteration of the loop a new "guess" of the squareroot is generated and this continues until the guess is a sufficiently close approximation of the squareroot.

Help with square root calculator

Hi: I am new to Java. Trying to write code for square root calculator of numbers 1-10. Have no idea where to start.

Using pencil and paper, manually extract the square root of a number, and write down the steps that you use to do that. Create a Java program that does those steps.
A Java Tutorial is here: http://java.sun.com/docs/books/tutorial/

Newtonian square root help! im getting a zero!

Hi guys, this is my first post here, adn I'm taking first level Java. I just recently wrote this code and at first it had 2 errors which i was finally able to debug, and made a clean compile. But now when i run it, it always gives me a zero output. Hope you can enlighten me. I know this code may seem elementary to some here, I'm a newbie, so I hope you guys understand! Thanks! Here's the code...
public class NewtonianSqRoot
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}

Hi !
Your logic is wrong.....and your code could definetly give the compiler errors... here is the error free code..
but, you are getting the same number....
import javax.swing.JOptionPane;
public class NewtonianSqRoot
public static void main (String[]args)
double num=0.00;
//System.out.println("The square root of " + number + " is " + result);
System.out.println("The square root is " + newtonian(num));
System.exit(0);
public static double newtonian(double number)
double nextGuess=0.00;
double lastGuess=0.00;
double tolerance = 0.0001;
String gradeString = JOptionPane.showInputDialog(null, "Please enter a number, \nthe program exits if the input is zero (0)", "Newtonian Square Root", JOptionPane.QUESTION_MESSAGE);
number = Double.parseDouble(gradeString);
lastGuess = number;
for (int count = 0; count < 20; count++)
nextGuess = ((number / lastGuess) + lastGuess) / 2.0;
if ((number - nextGuess) < tolerance)
return nextGuess;
else
nextGuess = lastGuess;
//System.exit(0);
return nextGuess;
}

ForLoop Square Root Calculator Program

* Programmer:      R McBride
* Date:           November 29, 2006
* Filename:     SquareRootForLoop.java
* Purpose:          This program uses for()loops to
*                    find the square root of a number
*                    up to four decimal places long.
import APCS.Keyboard;
class SquareRootForLoopTest1
     public static void main (String[]args)
          //define variables
          double SquareRoot = 0; //holds initial value
          double num1 = 0;
          boolean done = false;
          double tempReg1 = 0; //holds initial value of i
          double tempReg2 = 0; //holds
          double tempReg3 = 0;
          //while loop start
//          while (!done)
          //program input
          System.out.println("\t\t\tSQUARE ROOT CALCULATOR PROGRAM");
          System.out.println(""); //skip line
          System.out.println("Enter the number you would like to be square rooted or [0] to end the program.");
          System.out.print("");
          SquareRoot = Keyboard.readInt();
          System.out.println(""); //skip line
          System.out.println("You entered "+SquareRoot+".");
          System.out.println(""); //skip line
          //for loop function
          for(int i = 0; i < 999999; i++)
          tempReg1 = i * i;
          tempReg2 = i;
//          while (!done)
                    if(tempReg1 == SquareRoot)
                         System.out.println("The square root is "+tempReg2);
                         System.out.println("");
                    else
                         System.out.print("");
          //while loop check to finish
//          if(SquareRoot == 0)
//               done = true;
//          }//end while loop
}Here is my code for and of you who ever wants to use it. Our assignment was to make a square root calculator using ForLoop (a.k.a. no square root functions).
My question is, how would I make it increase by .0001 instead of 1 every loop? I need it to do that so I can find square roots of numbers of numbers that have decimals for squares.

* Programmer:      R McBride
* Date:           November 29, 2006
* Filename:     SquareRootForLoop.java
* Purpose:          This program uses for()loops to
*                    find the square root of a number
*                    up to four decimal places long.
import APCS.Keyboard;
class SquareRootForLoopTest1
     public static void main (String[]args)
          //define variables
          double SquareRoot = 0; //holds initial value
          double num1 = 0;
          boolean done = false;
          double tempReg1 = 0; //holds initial value of i
          double tempReg2 = 0; //holds
          double tempReg3 = 0;
          //while loop start
//          while (!done)
          //program input
          System.out.println("\t\t\tSQUARE ROOT CALCULATOR PROGRAM");
          System.out.println(""); //skip line
          System.out.println("Enter the number you would like to be square rooted or [0] to end the program.");
          System.out.print("");
          SquareRoot = Keyboard.readInt();
          System.out.println(""); //skip line
          System.out.println("You entered "+SquareRoot+".");
          System.out.println(""); //skip line
          //for loop function
          for(double i = 0; i < 999999; i+=.0001)
          tempReg1 = i * i;
          tempReg2 = i;
//          while (!done)
                    if(tempReg1 == SquareRoot)
                         System.out.println("The square root is "+tempReg2);
                         System.out.println("");
                    else
                         System.out.print("");
          //while loop check to finish
//          if(SquareRoot == 0)
//               done = true;
//          }//end while loop
}Still doesn't work. =[

Calculating Square root.

Hi, I'm suppose to write a programme that asks the user to enter a number. Find the square root of that number. Format theoutput to 4 decimal places. Use Math.sqrt(double a). i.e. to find the square root of 9.56, use: double sq = Math.sqrt(9.56);
This is what I have:
import java.io.*;
import java.math.*;
public class MethodEx2
static BufferedReader keyboard = new BufferedReader (new
InputStreamReader(System.in));
public static void main (String[] args) throws IOException
    double num = 0;
    System.out.print("Please enter a number");
    System.out.flush();
    num = Integer.parseInt(keyboard.readLine());
    calsqrt(num);
public static double sqrt (double number);
    double result = 0;
    result = sqrt(number);
    System.out.print(result);
}I keep getting an error saying that it cannot return a value from method whose result type is void at line 33, col 11.
How do I make the program to work?

Oh okay, thanks. I also found another typo. I fixed it but it still doesn't work. Anyway, this is my new code:
import java.io.*;
import java.math.*;
public class MethodEx2
static BufferedReader keyboard = new BufferedReader (new
InputStreamReader(System.in));
public static void main (String[] args) throws IOException
    double num = 0;
    System.out.print("Please enter a number");
    System.out.flush();
    num = Integer.parseInt(keyboard.readLine());
    sqrt(num);
public static double sqrt (double number)
    double result = 0;
    result = sqrt(number);
    System.out.print(result);
}It says that MethodEx2.java is missing return statement at line 17, column 3. I'm not sure which variable I should return. I tried 'number' and 'result' but something like this happens:
"C:\Program Files\JBuilder9\jdk1.4\bin\javaw" -classpath "E:\Lili\School\GR12 Comp Sci\Period 1\Programs\MethodExs\Ex2\Ex2\classes;C:\Program Files\JBuilder9\jdk1.4\demo\jfc\Java2D\Java2Demo.jar;C:\Program Files\JBuilder9\jdk1.4\demo\plugin\jfc\Java2D\Java2Demo.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\charsets.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\ext\dnsns.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\ext\ldapsec.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\ext\localedata.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\ext\sunjce_provider.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\im\indicim.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\jaws.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\jce.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\jsse.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\rt.jar;C:\Program Files\JBuilder9\jdk1.4\jre\lib\sunrsasign.jar;C:\Program Files\JBuilder9\jdk1.4\lib\dt.jar;C:\Program Files\JBuilder9\jdk1.4\lib\htmlconverter.jar;C:\Program Files\JBuilder9\jdk1.4\lib\tools.jar" MethodEx2
Please enter a number9
java.lang.StackOverflowError
     at MethodEx2.sqrt(MethodEx2.java:19)
     at MethodEx2.sqrt(MethodEx2.java:19)
     at MethodEx2.sqrt(MethodEx2.java:19)
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Square root formula

I have a program that has three columns, in the first column it lists numbers 0-10, in the second column I need to get the square root of each number in the first column. How would I get those results. What is the formula for the square root and how would I code it in? Here is my code so far
public class ThreeColumn {
    public static void main(String[] args) {
    System.out.println("Number \t Square root \t Even/Odd \t \n");
    int Ans1 = 0, Ans2 = 0, Ans3 = 0;
     for (int i = 0; i <= 10; i++)
System.out.println(Ans1 + "\t" );
Ans1++;
}Thanks

Here's something to play around with
import java.text.*;
class Testing
public Testing()
    DecimalFormat df = new DecimalFormat("0.00");
    String[] header = {"Number", "Sqr Root","Even/Odd"};
    String pad = "        ";
    for(int x = 0; x < header.length; x++) System.out.print(header[x]+" ");
    System.out.println("\n========================");
    String sqrNum;
    String[] evenOdd = {"Even","Odd"};
    for(int x = 1; x <= 10; x++)
      sqrNum = df.format(Math.sqrt(x));
      System.out.println((pad+x).substring(pad.length()-header[0].length()+(""+x).length())+
                         (pad+sqrNum).substring(pad.length()-header[1].length()+sqrNum.length()-1)+
                         (pad+evenOdd[x%2]).substring(pad.length()-header[2].length()+evenOdd[x%2].length()-1));
    System.exit(0);
public static void main(String[] args){new Testing();}
}

Square root calculation method

I'm trying to create java code that displays the square root of a number that the user enters, as I am not a great mathematician i cannot work out the logic for this. Any help would be great.
This is the part of my code that I need assistance with: n1 is the number that the user enters and n2 is where the square root is returned to. I'm guessing that i need to have a loop of some sort to divide n1, but am unsure of the conditions. The n1 is 0 so that after the total is calculated from n1, it is cleared for another number to be entered.
else if (source == btnSquareRoot)
n2 = ;
n1 = 0;
Please help!

An adequate method for finding a square root of a number x is this:
Choose a number that is too low to be the actual square root. (How you do this, I don't know, but zero is pretty small and might be low enough)
Call that number lo:
Choose a number that is too high to be the actual square root. (Same comment. well almost. Don't use zero for this - it's too small)
Call it hi:
Note, it is easy to check and see that the numbers you have choosen have the required properties. because lo*lo < x and hi*hi > x
Now you have two candidates for the square root, one of them too low and one of them two high. The actual square root lies somewhere between those two bounds.
What you want to do now is squeeze those bounds tighter. You do that by choosing a number that is between lo and hi. The mid point would be a nice choice. You may need to figure out how to compute that.
call that number a:
well now, either a was itself too low, too high or just right. If it was too low, why replace lo with a and you have just improved the lower bound, if it was too high...
surely you get the idea.
Now the question is: How long do you keep this up? Do you ever get the actual square root this way? The answer is: Of course you don't. Most of the time the actual square root requires an infinite number of decimals to represent it. All you are looking for is something that is good enough.
What does good enough mean? Did you want the number correct to 2 decimal places, to 4 decimals. What do you want? How can you tell if lo and hi are practically the same number?
If this was a homework problem, that of course goes back to what the teacher wants. If the teacher did not clearly specify, you need to go back and ask what they actually wanted.
On the other hand if you want to earn a reputations as a smart ass, you figure out some way to detect that the number that you are working on does not have an exact square root and when it does not, you simply print out the message "Sorry - the square root of the number you are looking for cannot be represented in a finite number of digits without resorting to the notation of continued fractions but here it is to 3 decimals..."
This is the sort of stuff that wins big bonus points with professors.
On the other hand the way you lose big points with the professors if you say something like that and CAN NOT explain to him what a continued fraction is and explain what you meant by the qualification.
The other thing that loses big is to cut a chunk of text directly off of a web page and try to fob it off as your own program. Since most professors know how to google, they will stuff some unlikely looking phrase like "notation of continued fractions" and boom they are immediately at this page.
My point is that if you are going to be a smart ass, you MUST do it carefully or you lose all credibility.
Sorry for the digression there. Hope this algorithm outline is enough to get you thinking.
Wow! Killer. I previewed my question and it put asterisks in where I wrote the word, "ass". That is Fucking awesome! They've put in some kind of bad word filter so that when I get fucking abusive with my language and call you a no good shit for brains mother fucker. it cleans it all up for me. That is just too fucking sweet! It sure is a load off of my mind. I guess they just couldn't figure out how to put in !@%#^ which is the way that you actually are supposed to replace vulgarity in the printed word. Oh well, we always knew that this site was maintained by a bunch of **** ***** * **** ***** ** ***** * * *********!

Square root is not working...

I wrote a simple program to derermine the square root of a number, but its not working: Heres the code:
class root{
     static public void main(String[] args){
          int square = Math.sqrt[4];
          System.out.println("the square root is " + square);
}I get this error mesage when running:
C:\java_apps>javac root.java
root.java:3: cannot find symbol
symbol : variable sqrt
location: class java.lang.Math
int square = Math.sqrt[4];
^ (arrow points to dot after "Math")
1 error
Thanks!
Jake

Math.sqrt() is a method, so you have to invoke it with parentheses:int square = Math.sqrt(4);
> root.java:3: cannot find symbol
symbol : variable sqrt
location: class java.lang.Math
int square = Math.sqrt[4];
^ (arrow points to dot after "Math")Because of the square bracket the compiler is looking for an array sqrt inthe Math class. It can't find one and so you get the message.
(Note that the Math static methods tend to return double not int, so you will have to
deal with that as well.)

Fast Inverse Square Root

I expect no replies to this thread - because there are no
answers, but I want to raise awareness of a faculty of other
languages that is missing in Flash that would really help 3D and
games to be built in Flash.
Below is an optimisation of the Quake 3 inverse square root
hack. What does it do? Well in games and 3D we use a lot of vector
math and that involves calculating normals. To calculate a normal
you divide a vector's parameters by it's length, the length you
obtain by pythagoras theorem. But of course division is slow - if
only there was a way we could get 1.0/Math.sqrt so we could just
multiply the vector and speed it up.
Which is what the code below does in Java / Processing. It
runs at the same speed as Math.sqrt, but for not having to divide,
that's still a massive speed increase.
But we can't do this in Flash because there isn't a way to
convert a Number/float into its integer-bits representation. Please
could everyone whinge at Adobe about this and give us access to a
very powerful tool. Even the guys working on Papervision are having
trouble with this issue.

that's just an implementation of newton's method for finding
the zeros of a differentiable function. for a given x whose inverse
sq rt you want to find, the function is:
f(y) = 1/(y*y) - x;
1. you can find the positive zero of f using newton's method.
2. you only need to consider values of x between 1 and 10
because you can rewrite x = 10^^E * m, where 1<=m<10.
3. the inverseRt(x) = 10^^(-E/2) * inverseRt(m)
4. you don't have to divide E by 2. you can use bitwise shift
to the right by 1.
5. you don't have to multiply 10^^(-E/2) by inverseRt(m): you
can use a decimal shift of inverseRt(m);
6. your left to find the positive zero of f(y) = 1/(y*y) - m,
1<=m<10.
and at this point i realized what, i believe, is a much
faster way to find inverse roots: use a look-up table.
you only need a table of inverse roots for numbers m,
1<m<=10.
for a given x = 10^^E*m = 10^^(e/2) *10^^(E-e/2)*m, where e
is the largest even integer less than or equal to E (if E is
positive, e is the greatest even integer less than or equal to E,
if E is negative), you need to look-up, at most, two inverse roots,
perform one multiplication and one decimal shift:
inverseRt(x) = 10^^(-e) * inverseRt(10) *inverseRt(m), if
E-e/2 = 1 and
inverseRt(x) = 10^^(-e) * inverseRt(m), if E-e/2 = 0.

Sampling procedure-Square root n+1

Hi All,
I need to calculate the sample size based on the Containers & Square root n+1.
How to configure the system & how to maintain the master data to get the sample size.
User also insisted only the sample size has to be calculated, but he will enter only one composite result entry
Regards
Subbu

Hi Subbu,
you can use Physical Sample to control the sample Size.
You must to define a Sample Drawing Procedure (QPV2) as relevant for Container Number, then assign the formula 'Square Root n1´ ( TRUNC(SQRT(P2)1) ) in the partial sample. In this scenario, I suggest you to configure the "Container number' as a mandatory field during the goods receipt once the system will need this information to calculate the sample. For a single result recording, you can define a pooled sample.
I hope it can help you.
Best regards,
Robson

For square root of biginteger

Hi everyone,
would you please tell me the link to get the implementation for one BigInteger's squre root?
Regards,
Ping

Although there are no methods to get the square root in the API, you can go through the recursive solution using basic math operators:
An = (Al / X + Al)/2 where Al is the last number from the resursion (or initially a guess), X is the number you are trying to find the root of, and An is your next closest answer. By recursively feeding An into Al, each iteration will result in An being closer to the actual square root. Repeat until you are within the error of margin you want. However, for this process to work, you need to have floating point values, so you may have to convert the values to BigDecimal objects first. I haven't been able to test this, so I hope it works.
-JBoeing

Calculator square root event listener

I am trying to write a calculator using JButtons added onto Jpanels. Getting numbers to add , divide, multilpy etc. is simple enough, but i need to write an event handler for a button which calculates the square root of a given number. This is easy enough if you want the square root of 4,9,16,25 etc but if you enter a number which produces a decimal point value it doesn't work.
I am trying to do this using a variable "number" which is of type double, but have tried using it as a float, but it still doesn't work! here is my code for handling a click on the Square root JButton.
     if(e.getActionCommand().equals("Sqrt")){
     number = new Double(jtf.getText().trim()).doubleValue();
     double rootNum = 1.000;
     while(rootNum*rootNum <= number){
          double rootTester = rootNum * rootNum;
          if(rootTester == number){
               jtf.setText(String.valueOf(rootNum));
rootNum = rootNum+1;

How about
double answer = Math.sqrt(number);

Fastest square root algorithm

I was looking for a fast algorythm for integer square roots and I found this one http://medialab.freaknet.org/martin/src/sqrt/.
The algorithm comes from a book by Mr C. Woo on how to do maths on an abacus.
I post the javaized version here in case anyone finds it interesting.
I believe this is the fastest square root function in existance for integers (and in game programming much of the time you aren't interested in fractions)
/* Fast interger square root adapted from algorithm by Martin Guy @ UKC, June 1985.
    *   Origonally from a book on programming abaci by Mr C. Woo.
   public static int fastSqrt2(int n)
      int op, res, one;
      op = n;
      res = 0;
      /* "one" starts at the highest power of four <= than the argument. */
      one = 1 << 30;   /* second-to-top bit set */
      while (one > op) one >>= 2;
      while (one != 0)
         if (op >= res + one)
            op = op - (res + one);
            res = res + (one<<1);
         res >>= 1;
         one >>= 2;
      return(res);
   }

public static double sqrt(double a){
if(a<0) throw new IllegalArgumentException("number<0");
double precision=0.001;
double x_nMinus1 = -1;
double x_n = 1;
while( Math.abs(x_n - x_nMinus1) > precision ) {
    x_nMinus1 = x_n;
    x_n = (x_nMinus1 * x_nMinus1 + a) / (2*x_nMinus1);
return x_n;
}

Square Root of a number

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